Problem: Solve for $t$, $ \dfrac{1}{16t^3} = \dfrac{t + 1}{4t^3} - \dfrac{5}{4t^3} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16t^3$ $4t^3$ and $4t^3$ The common denominator is $16t^3$ The denominator of the first term is already $16t^3$ , so we don't need to change it. To get $16t^3$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{t + 1}{4t^3} \times \dfrac{4}{4} = \dfrac{4t + 4}{16t^3} $ To get $16t^3$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{5}{4t^3} \times \dfrac{4}{4} = -\dfrac{20}{16t^3} $ This give us: $ \dfrac{1}{16t^3} = \dfrac{4t + 4}{16t^3} - \dfrac{20}{16t^3} $ If we multiply both sides of the equation by $16t^3$ , we get: $ 1 = 4t + 4 - 20$ $ 1 = 4t - 16$ $ 17 = 4t $ $ t = \dfrac{17}{4}$